Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.4 - The Cross Product - Exercises 12.4 - Page 720: 48

Answer

$5$

Work Step by Step

$ \overrightarrow{AB} = \langle 1-0, 2-0,0-0 \rangle= \langle 1,2, 0 \rangle$ $ \overrightarrow{AC} = \langle 0, -3, 2 \rangle$ $ \overrightarrow{AD} = \langle 3, -4, 5 \rangle$ The volume of the parallelopiped is the absolute value of the triple scalar product, $( \overrightarrow{AB}\times \overrightarrow{AC})\cdot \overrightarrow{AD}=\left|\begin{array}{rrr} 1 & 2 & 0\\ 0 & -3 & 2\\ 3 & -4 & 5 \end{array}\right|=1(-15+8)-2(0-6)+0$ $=-7+12=5$ $V=|5|=5$
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